在我自己的shell中创建后台进程

Question

I am writing my own unix shell as a part of my assignment and I couldn't handle creating background processes. I wrote a signal-handling function and implemented the necessary(from my point of view-but obviously not good enough) parts to my code as follows :

void handler(int sig)
{
    int pid;
    int status;
    pid = wait(NULL);
    printf("[%d]retval: %d \n", pid, WEXITSTATUS(status));
    fflush(stdout);
}

int main() { 
.....
....
struct sigaction sigchild; 
memset (&sigchild, 0, sizeof(sigchild)); 
sigchild.sa_handler = handler;
sigchild.sa_flags = SA_SIGINFO | SA_NOCLDWAIT;
...
...
if(isBackground)  //background process
{
    sigaction(SIGCHLD, &sigchild, 0);
}

When I enter "sleep 5 &" for example and then enter "ps" to see the processes there is no sleep. What may be the problem? Thanks in advance.

Answer 1

The child is the background process. The parent is the foreground process. SIGCHLD is a signal sent to a parent when the child terminates. Change the if (isbackground to if(!background)